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aleksley [76]
3 years ago
13

Two pressure gauges measure a pressure drop of 16.3 psi (lb/in.2) at the entrance and exit of an old buried pipeline. The origin

al drawings have been lost. If the 6-in. galvanized iron pipe carries water at 68°F with a flow rate of 1.64 cfs (ft3/s), determine the length of the horizontal underground line ignoring minor losses.
Engineering
1 answer:
Sholpan [36]3 years ago
7 0

Answer: 866.25ft

Explanation:

Pressure drop= 16.3 psi

1 feet of water column= 0.4335psi

x= 16.3psi

Therefore, head loss of water= 16.3/0.4335=37.6ft

In inches, head loss of water= 37.6 × 12=451.2 inches

Q= 1.64ft^2/s

A= pi/4 D^2

Where D= 6 inches, 0.5ft

A= 3.142/4 × (0.5)^2

A= 0.1962ft^2

V= Q/A= 1.64/ 0.1962

V= 8.35ft/s, 100.28 in/s

REGNOLD'S NO

Re= SvD/ μ=

Where, D= 6 inches, V= 100.28 inches/s, S=62.4lb/ft^3= 0.03611lb/inches^3

At 68F, Dynamic viscosity μ= 1.0016× 10^-3NS/m^2

1psi= 6895N/m^3

μ= 1.0016× 10^-3NS/m^2/6895N/m^3

μ= 1.452× 10^-7lbs/ins^2

Re= 0.03611× 100. 28×6/1.42×10^-7

Re= 1.49×10^8

e/D----- Relative roughness

e= 0.0005ft

e/D= 0.001

Therefore, for Re= 1.49×10^8 and e/D= 0.001

F=0.02

Head loss is given by Dancy-Weisbach formula

hL= fLV^2/2gD

Wherr g= 9.81m/s^2= 386.12inches/s^2

451.2= 0.02×L×(100.28^2)/2 × 386.12inches/s^2× 6

L= 10395inches

L= 10395/12= 866.25ft

Therefore, the length of pipe is 866.25ft.

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Answer:

a) Check explanation for this

b)Rate law is  Rate = \frac{k_{1}k_{4}  }{k_{3}+ 2k_{4}  } [H_{2} ]

c) The rate does not depend on the concentration of CO₂

Explanation:

a) Elementary steps for the RWGS reaction:

  • Dissociative adsorption of the H₂ Molecule

                 H_{2} $\xrightarrow{\text{k1}}$H + H   (Fast process)

  • Reversible Reaction between CO₂ and H

                \[ CO_{2} + H\mathrel{\mathop{\rightleftarrows}^{\mathrm{k2}}_{\mathrm{k3}}}COOH \] (Fast Process)

  • Slow dissociation of COOH into gaseous CO and absorbed OH

                COOH $\xrightarrow{\text{k1}}$ CO + OH (Slow process)

  • Fast hydrogenation of the OH to form H₂O

                   OH + H $\xrightarrow{\text{k5}}$H_{2} O (Fast process)

b) Derivation of the rate law

We need to determine the rate law for H, OH and COOH because these are the intermediates for this reaction.

The steady state approximation is applied to a consecutive reaction with a slow first step and a fast second step (k1≪k2). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product.

Rate of consumption = Rate of production

For COOH:

Using steady state approximation

\frac{d[COOH]}{dt} = 0

k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]\\

[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4}  } \\

For H:

\frac{d[H]}{dt} = 0

k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]

[H]= \frac{k_{1}[H_{2}]  }{k_{5}[OH] +k_{2}[CO_{2}]}\\

For OH:

\frac{d[OH]}{dt} = 0

k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}\\

The rate of the overall reaction is determined by the slowest step of the reaction. The slowest process is the dissociation of COOH

Therefore the overall rate of reaction is:

Rate = k_{4} [COOH]\\

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Simplifying the equation above, the rate law becomes

Rate = \frac{k_{1}k_{4}  }{k_{3}+ 2k_{4}  } [H_{2} ]

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Answer:

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Hence the average flow rate of the air conditioning unit of this room

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Average flow rate = V/ t

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